Investigating Hard Inequalities: Understanding Part (f)

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If (f) is the part you are having trouble with, then presumably you have already proved that 2x_n^2- (2n-1)x- (n+1)= 0 (part (e)). Now you want to find the smallest n such that x_n< n+ 0.05. You could, for example, solve that using the quadrative formula and compare the solutions to n+ 0.05. Have you calculated some values of x_n? What are x_0 x_1, etc.?
 
Thanks for the help. I am still confused as to how the markscheme answers have come about which I attached above.

Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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